Friday, 26 December 2014

integration - An integral involving error functions and a Gaussian

Let d1 be an integer and let A:={Ai}di=1 be real numbers. We consider a following integral:
I(d)(A):=0eu2[di=1erf(Aiu)]du


By expanding the error functions in Taylor series and then integrating term by term we found the answer for d=1 and d=2. We have:

πI(d)(A)={arctan(A1)if d=1arctan(A1A21+A21+A22)if d=2

Now the question is how do we derive the result for arbitrary values of d?

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